424 RESTRAINED TORSION FLEXURE OF PRISMATIC MEMBER CHAP 13 Hint One can write if fi3 2 I3 X2 V 2 r X3 V2rldA 2 02 2 32 I3 14 Also show that 13 2 2 13 12 Specialize Equations 13 84 and 13 85 for the case where the cross section is symmetrical with respect to the X 2 axis Utilize JjHe X2 x3 HNo 2 x 3 dA 0 where He is an even function and Ho an odd function of x3 Evaluate the co efficients for the channel section of Example 13 5 Finally specialize the equations for a doubly symmetric section 13 13 Specialize 13 88 for a doubly symmetrical cross section Then specialize further for negligible transverse shear deformation due to flexure and warping The symmetry reductions are X2 3 P2 3 X C2r 0 3 t 14 1 X3r 0 1 The centroidal axis is a plane curve 2 The plane containing the centroidal axis also contains one of the principal inertia axes for the cross section 3 The shear center axis coincides with or is parallel to the centroidal axis However the present discussion will be limited to the case where the shear center axis lies in the plane containing the centroidal axis i 17i 0 13 14 Consider the two following problems involving doubly symmetric cross section a Establish linearized incremental equations by operating on 13 88 and retaining only linear ternms in the displacement increments Specialize for a doubly symmetric cross section see Prob 13 12 b Consider the case where the cross section is doubly symmetric and the initial state is pure compression F t Determine the critical load with respect to torsional buckling for the following boundary conditions 1 o 2 o 1 f 0 df dX 0 0 at unrestrained warping L Neutral equilibrium buckling is defined as the existence of a nontrivial solution of the linearized incremental equations for the same external load One sets U3 01 02 X2 i j Yl F1 P t2 We consider the centroidal axis to be defined with respect to a global reference frame having directions X1 and X2 This is shown in Fig 14 1 The orthogonal unit vectors defining the orientation of the local fame Y Y2 at a point are denoted by Ft 2 where t points in the positive tangent direction and t x t2 t3 Item 2 requires Y2 to be a principal inertia axis for the cross section restrained warping at x 0 L INTRODUCTION GEOMETRICAL RELATIONS A member is said to be planar if 1 A 2 3 0 0 72 173 i 1 1I4 Planar Deformation of a Planar Member 03 f and determines the value of P for which a nontrivial solution which satisfies the boundary conditions is possible Employ the notation introduced in Example 13 7 13 15 Determine the form of V the strain energy density function strain energy per unit length along the centroidal axis expressed in terms of displace ments Assume no initial strain but allow for geometric nonlinearity Note that V V when there is no initial strain A I 12 I4 l1 I Fig 14 1 Geometrical notation for plane curve 425 426 PLANAR DEFORMATION OF A PLANAR MEMBER CHAP 14 By definition t dr t dS dxl dS dx 2 dx2 dS dS 2 1 3 t dS 1 dxi I 14 7 R o Jo v 1 a03 dt2 dS R t2 14 3 1 R tl where 1 R dt l dd2 d S t x2 d2 x 2 dxl dS dS dS According to this definition R is negative when dil dS points in the negative direction e g for segment AB in Fig 14 1 One could take 2 ii the unit normal vector defined by t2 I dtl dt dS 14 4 x t2 3 but this choice is inconvenient when there is a reversal in curvature Also this definition degenerates at an inflection point i e when dF dS If the sense of the curvature is constant one can always orient the X 1 X 2 frame so that 2 coincides with hi to avoid working with a negative R To complete the geometrical treatment we consider the general parametric representation for the curve defining the centroidal axis X d2x d2 d22 d d d 2 d dxl A planar member subjected to in plane forces Xi X plane our notation will experience only in plane deformation In what 2follows for we develop the governing equations for planar deformation of a arbitrary member This formulation is restricted to the linear geometric case planar The two basic solution procedures namely the displacement and force methods are described and applied to a circular member We also present a simplified formulation Marguerre s equations which is valid for a shallow member Finally we include a discussion of numerical integration techniques since one must resort to numerical integration when the cross section is not constant 14 2 FORCE EQUILIBRIUM EQUATIONS The notation associated with a positive normal cross dCSI rather than according to t V d x2 a 14 2 12 The differentiation formulas for the unit vectors are dS d 2 a kTT it follows that dxl 1 427 2 and 1 R in terms of v are 14 1 dSS 12 Since we are taking t2 according to t1 x t 2 FORCE EQUILIBRIUM EQUATInm n increasing y Using 14 6 the expressions for ti ldS dS SEC 14 2 l Y i e a cross section whose outward normal points in the S direction issection shownin Fig 14 2 We use the same notation as for the prismatic case except that now the vector 4 Y3 Y t1 J1 x2 x 2 Y where y is a parameter The differential arc length is related to dy by dS dX 2 12 dv dy 14 6 According to this definition the S sense coincides with the direction of t We summarize here for convenience the essential geometric relations for a plane curve which are developed in Chapter 4 Fig 14 2 Force and moment components acting on a positive cross section PLANAR DEFORMATION OF A PLANAR MEMBER 428 CHAP 14 SEC 14 3 components are with respect to the local frame Y1 Y 2 Y 3 rathe r than the basic frame X1 X 2 X 3 The cross sectional properties are defineed by A ff dy dY dy 13 ff y2 2 dA dA 2 12 f y3 b bl l b 212 Introducing the component expansionsin 14 12 and using the differentiation formulas for the unit vectors 14 3 lead to the following scalar differential equilibrium equations 14 10 r F i F 72 F F 14 11 1 R t b 0 14 14 Note that the force equilibrium equations are coupled due to the curvature The moment equilibrium equation has the saime form as for the prismatic case M3 Note that 3 is constantfor a planar member I t F2 dS dF 2 F1 dS R dM F2 m O dS In this case we work with reduced expressions for F and M see Fig 14 3 and …